Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations $x=x(t)$ and $y=y(t)$, then $\frac{dy}{dx} = \frac{y’(t)}{x’(t)}$

• This can also be applied to second-order derivatives
• Ides is to “cut an area” around the point where we want to take a derivative to make it a function, since parametric curves do not pass the vertical line test
• To make it easier to graph, we can find the horizontal tangent $\frac{dy}{dt} = 0$, and the vertical tangent $\frac{dx}{dt}=0$
  • We can also find the x and y intercepts

Area under a Parametric Curve

Area under a curve where $a≤t≤b$ is given by $A = \int _a ^b y(t)x’(t)dt$

• In polar coordinates, $A=\frac{1}{2} \int _{\alpha} ^{\beta} r^2 d\theta$

Surface Area of a Parametric Curve

$S = 2\pi \int _a ^b y(t) \sqrt{x’(t)^2+y’(t)^2}dt$

Area of Revolution